Solution
= Solution
Set $\omega=\pi/(2\tau)$. The condition $r\tau=-\pi/2$ says $r=-\omega$. Then
$$
n(t)=C\cos(\omega t)
$$
satisfies
$$
n'(t)=-C\omega\sin(\omega t),
\qquad
r n(t-\tau)=-C\omega\cos(\omega t-\pi/2)
=-C\omega\sin(\omega t).
$$
It is periodic with period $2\pi/\omega=4\tau$. Every nonzero such solution is negative during half of each period, so it cannot represent a population size.